In the AP (12,17,22,\ldots), find the sum from the (21)st term to the (40)th term.

(3190)

In an arithmetic progression, each term changes by the same common difference. Here the first term is 12 and the common difference is 5, so the term at position n is found from the rule \(a_n=12+(n-1)5\). The requested terms run from the 21st through the 40th, giving 20 terms in all. Their first term is \(a_{21}=112\), and their last term is \(a_{40}=207\). The sum of an AP can be found by multiplying the number of terms by the average of the first and last terms.

Thus, the required sum is \(20\times\frac{112+207}{2}=10\times319=3190\). Equivalently, one may calculate \(S_{40}-S_{20}\), because subtracting the first 20 terms leaves terms 21 to 40. Therefore option A is correct. A common error is to subtract \(S_{21}\), which would remove the 21st term too and give the wrong range.

This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the sum of the first $n$ terms of an AP.